$\lim_{x\to -3}\dfrac{x^3+x^2-6x}{x^2+3x}=$
Answer: Substituting $x=-3$ into $\dfrac{x^3+x^2-6x}{x^2+3x}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{x^3+x^2-6x}{x^2+3x}$ can be simplified as $x-2$, for $x\neq 0,-3$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $0$ and $-3$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x^3+x^2-6x}{x^2+3x}=x-2$ for all $x$ -values in the interval $(-4,-2)$ except for $x=-3$. Therefore, $\lim_{x\to -3}\dfrac{x^3+x^2-6x}{x^2+3x}=\lim_{x\to -3}(x-2)=-5$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -3}\dfrac{x^3+x^2-6x}{x^2+3x}=-5$.